What is the volume of CO_2 in liters produced | vedclass

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(B) The chemical equation for the decomposition of $BaCO_3$ is: $BaCO_3(s) 	o BaO(s) + CO_2(g)$ First,calculate the molar mass of $BaCO_3$: $M = 137

Vedclass · Accepted Answer

$1.12$

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(B) The chemical equation for the decomposition of $BaCO_3$ is: $BaCO_3(s) 	o BaO(s) + CO_2(g)$ First,calculate the molar mass of $BaCO_3$: $M = 137 + 12 + (3 	imes 16) = 197 \ g/mol$ Calculate the number of moles of $BaCO_3$: $n = \frac{9.85 \ g}{197 \ g/mol} = 0.05 \ mol$ According to the stoichiometry,$1 \ mol$ of $BaCO_3$ produces $1 \ mol$ of $CO_2$. Therefore,$0.05 \ mol$ of $BaCO_3$ produces $0.05 \ mol$ of $CO_2$. At $STP$,the volume of $1 \ mol$ of gas is $22.4 \ L$. Volume of $CO_2 = 0.05 \ mol 	imes 22.4 \ L/mol = 1.12 \ L$.

What is the volume of $CO_2$ in liters produced at $STP$ by the complete decomposition of $9.85 \ g$ of $BaCO_3$? (Atomic mass of $Ba = 137$,$C = 12$,$O = 16$)

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